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Substituting the above expression into Eq. 422, we find the maximum height to be
2
22
2
00
0
0
max
0
0
max
max
0
0
sin
sin
sin
11
(s
i
n)
s
i
n
.
2
vv
v
yv
t
g
t
v
g
gg
g
θθ
θ
§·
=−
=
−
=
¨¸
©¹
To find the time when the player is at
max
/2
yy
=
, we solve the quadratic equation given
in Eq. 422:
2
0
0
max
0
0
sin
(2
2)
sin
i
.
24
2
2
v
tg
t
t
±
±
==
=
−
¡
=
With
tt
−
=
(for ascending), the amount of time the player spends at a height
max
≥
is
m
a
x
max
max
sin
(2
sin
sin
1
0.707
2
2
v
t
t
t
t
g
−
−
Δ
Δ=
− =
−
=
=
¡
.
Therefore, the player spends about 70.7% of the time in the upper half of the jump. Note
that the ratio
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics

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